Let $V$ be a bounded open set in $\mathbb{R}^{n}$ with $n\geq2$, and $u$ a subharmonic function on an open set containing the closure $ \overline{V} $ of $V$. Suppose we have $u<M$ on $V$ ($M$ is a constant). Can we conclude that $u\leq M$ on $ \overline{V} $?

  • $\begingroup$ Do you allow your subharmonic functions to attain the value $-\infty$? $\endgroup$ – Daniel Fischer Jul 21 '17 at 15:07
  • $\begingroup$ Yes, but cannot be identically $-\infty$. $\endgroup$ – M. Rahmat Jul 22 '17 at 0:52

I have a partial answer to this question: if $u$ is continuous at each point of the boundary $\partial V$ of $V$, the answer is yes! In fact, we have by continuity of $u$ and the maximum principle, $$u(\zeta)=\limsup_{\substack{x\rightarrow\zeta\\(x\in V)}}u(x)\leq M$$ for all point $\zeta\in\partial V$, which proves the claim!

  • $\begingroup$ This actually works for all functions that are continuous on the boundary, as $u(x)<M$ for all $x\in V$ by assumption. No need for the maximum principle. $\endgroup$ – Giuseppe Negro Jul 27 '17 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.